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Mass of Solute Needed
58.44
grams
Moles required 1 mol

What This Calculator Does

The Mass Needed for Molarity Calculator tells you exactly how many grams of a solute you must weigh out to prepare a solution of a desired concentration (molarity) and volume. It is an essential bench tool for chemistry students, lab technicians, and researchers who prepare standard solutions.

How to Use It

Enter three values: the desired molarity in moles per liter (mol/L), the final volume of solution in liters, and the molar mass of your compound in grams per mole (g/mol). The calculator returns the mass of solute in grams, along with the total moles required.

The Formula Explained

The calculation is based on the definition of molarity, \(M = \text{moles} \div \text{volume}\). Rearranging gives \(\text{moles} = M \times V\). Since \(\text{mass} = \text{moles} \times \text{molar mass}\), we combine these into a single step:

$$\text{mass (g)} = M \times V\text{(L)} \times MW$$

Molarity must be in mol/L, volume in liters, and molar mass in g/mol so that the units cancel to leave grams.

Diagram showing molarity times volume times molar mass equals mass in grams
Mass equals molarity multiplied by volume in liters multiplied by molar mass.

Worked Example

Suppose you want to make 0.5 liters of a 2 mol/L sodium chloride (NaCl) solution. NaCl has a molar mass of about 58.44 g/mol. The mass needed is:

$$\text{mass} = 2 \times 0.5 \times 58.44 = \mathbf{58.44 \text{ g}}$$ You would also need \(2 \times 0.5 = 1\) mole of NaCl.

Volumetric flask, weighed solute powder, and a balance scale
Weigh the calculated solute mass and dissolve it to the target volume.

FAQ

What if my volume is in milliliters? Convert to liters first by dividing by 1000 (e.g. 250 mL = 0.25 L).

Where do I find the molar mass? Add up the atomic masses of all atoms in the chemical formula, or look it up on a periodic table or chemical supplier datasheet.

Does this account for hydrates or purity? No. If your reagent is a hydrate, use the hydrated molar mass. For reagents below 100% purity, divide the result by the fractional purity.

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